Okay, let's try. Consider the following fact: $$$dp[i][j][k]$$$ is how many valid sequences exist, such that their length is $$$i$$$, last number is $$$j$$$, and it has been repeated $$$k$$$ times in the suffix.

First off, if $$$k > A[j]$$$, i.e. for example we're looking at $$$dp[12][4][3]$$$ from your input. Now, how many such valid sequences exist? Obviously 0, since by our input we cannot write more than two consecutive "4"-s (and moreover, our $$$N = 2$$$).

Okay, for the sake of convenience, let's now consider some other input that allows taking bigger sequences of equal numbers and asks for greater length sequences, so that $$$dp[12][4][3]$$$ is valid.

Secondly, how can we get that $$$dp[12][4][3]$$$? Remember, it means the number of sequences of length 12 that end in exactly three "4"-s. But hey, every such sequence is made in the following fashion: we take a sequence of 11 elements with exactly two consecutive "4"-s in the end and append one "4" to the end! So there are as many $$$dp[12][4][3]$$$-s as there are $$$dp[11][4][2]$$$-s! Observation 1: $$$dp[i][j][k]$$$ = $$$dp[i - 1][j][k - 1]$$$ for valid combinations of $$$j$$$ and $$$k$$$, except...

We forgot to take one important case into consideration, though. What if our last number is repeated only one time in the end, i.e. $$$dp[i][j][1]$$$? How can we get all such sequences? Well, we could append $$$j$$$ to any sequence of length $$$i - 1$$$, except for those that end with $$$j$$$ — because then we would have more than one repeating last symbol! Observation 2: $$$dp[i][j][1]$$$ = sum over all $$$dp[i - 1][m][n]$$$, where $$$m \ne j$$$.

Now try to combine these two observations into a correct algorithm. I might have missed a couple of things, but on the whole I think it is right.

#include <bits/stdc++.h>
using namespace std;

int dp[100005][7][51] = {};
int a[7];

int main() {
    int n;
    cin >> n;
    for (int i = 1; i <= 6; i++) cin >> a[i];
    for (int i = 1; i <= 6; i++) dp[1][i][1] = 1;

    for (int i = 2; i <= n; i++) {
        for (int j = 1; j <= 6; j++) {
            for (int p = 1; p <= 6; p++) {
                if (p == j) continue;
                for (int k = 1; k <= a[p]; k++) dp[i][j][1] += dp[i - 1][p][k];
            }
            for (int k = 2; k <= a[j]; k++) {
                dp[i][j][k] = dp[i - 1][j][k - 1];
            }
        }
    }
    int ans = 0;
    for (int i = 1; i <= 6; i++) {
        for (int j = 1; j <= a[i]; j++) ans += dp[n][i][j];
    }
    cout << ans;
}